Is the Rindler vacuum invariant under Poincare symmetries? More generally, when we quantize fields in the Rindler space and obtain the Fock space of Rindler particles - does that carry a unitary representation of the Poincare symmetries? It should not, because Rindler spacetime is not invariant under global translations. At the same time the Killing vectors in Rindler space are well defined except at the horizon.
Is there some subtlety related to this?
 A: It depends on the requirements  you impose on such a representation. So we assume that there is a strongly continuous unitary representation ${\cal P} \ni p \mapsto U_p$ of the orthochronous proper Poincaré group working on the Fock space constructed upon the   Rindler vacuum $\Omega_R$ exploiting the standard  static space quantization (the notion of time being the boost one) and  we assume that $$U_p\Omega_R= \Omega_R\quad \forall p \in \cal P\:.$$
In principle this is possible. What it is not possible is that the action of that $U$ on Rindler fields $\phi(f)$ (constructed out the standard procedure with creation of annihilation operators referred to $\Omega_R$) smeared with smooth functions $f$  supported in $\cal W_R$, also implements $\cal P$ on the algebra of fields into the natural geometric way:
$$U_p\phi(f) U_p^* = \phi(f\circ p^{-1})\:.$$ 
This is impossible because $p(\cal W_R) \not \subset \cal W_R$ for some $p \in \cal P$. It is possible if restricting $\cal P$ to the three dimensional Lie subgroup generated by the boost along $x$ and by the spatial translations along $y$ and $z$.  
